This question tests AP Physics C understanding of Ampère's Law, specifically its application to calculate the magnetic field inside a toroidal inductor. Ampère's Law is ideal for toroids due to their circular symmetry, where the magnetic field lines are circles centered on the toroid axis and confined to the core region between inner and outer radii. In this scenario, the toroid has 300 turns carrying 2.0 A, with inner radius 4.0 cm and outer radius 9.0 cm, and we evaluate the field at r = 5.0 cm using a circular Amperian loop that links all turns. Choice C is correct because applying Ampère's Law gives B(2πr) = μ₀NI, so B = μ₀NI/(2πr) = (4π×10⁻⁷)(300)(2.0)/(2π×0.05) = 1.5×10⁻³ T. Choice A would result from using a different radius, while choices B and D represent various calculation errors. To help students: Stress that the magnetic field in a toroid varies inversely with radius within the core region. Practice setting up Ampère's Law for different toroidal geometries and emphasize that the enclosed current includes contributions from all N turns when the loop is inside the toroid.

This question tests AP Physics C understanding of Ampère's Law, specifically its application to find the magnetic field inside a long solenoid. Ampère's Law is ideal for solenoids because the interior field is uniform and axial while the exterior field is negligible, allowing us to use a rectangular Amperian loop with one side inside and one outside. In this scenario, the solenoid has turn density n = 1200 m⁻¹ and carries current I = 0.80 A, with the rectangular loop having length ℓ inside where the field contributes to the line integral. Choice B is correct because applying Ampère's Law gives Bℓ = μ₀(nℓ)I, which simplifies to B = μ₀nI = (4π×10⁻⁷)(1200)(0.80) = 9.6×10⁻⁴ T. Choice A would result from using 1.0 A instead of 0.80 A, while choice D might come from calculation errors or unit confusion. To help students: Stress that the magnetic field inside an ideal solenoid depends only on μ₀, n, and I, not on the solenoid's radius or the specific loop chosen. Practice recognizing when approximations (like neglecting fringing fields) are valid and ensure proper unit handling with turn density.

This question tests AP Physics C understanding of Ampère's Law, specifically its application to find the magnetic field in a coaxial cable between the inner and outer conductors. Ampère's Law is ideal for this geometry due to cylindrical symmetry, where the magnetic field forms circles around the central axis and has magnitude depending only on radius. In this scenario, the inner conductor (radius 1.0 mm) carries 6.0 A forward while the outer conductor (at radius 5.0 mm) carries 6.0 A return current, and we calculate the field at r = 3.0 mm where the Amperian loop encloses only the inner conductor's current. Choice D is correct because applying Ampère's Law gives B(2πr) = μ₀I, so B = μ₀I/(2πr) = (4π×10⁻⁷)(6.0)/(2π×0.003) = 2.0×10⁻⁴ T. Choice C (zero) would apply outside the outer conductor where both currents cancel, not between the conductors. To help students: Stress the importance of identifying which currents are enclosed at different radial positions. Practice sketching the field configuration in coaxial cables and understanding why the field is zero outside when currents are equal and opposite.

Applying Ampère's law: $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I$$. For a circular path, $$B$$ is constant and tangent to the circle, so $$B(2\pi r) = \mu_0 I$$, giving $$B = \frac{\mu_0 I}{2\pi r}$$.

The key to successfully applying Ampère's law is choosing an Amperian loop that takes advantage of the symmetry of the problem, allowing the magnetic field to be constant along portions of the path and making the integral tractable.

Ampère's law is most useful when the current configuration has sufficient symmetry that the magnetic field has constant magnitude along portions of the chosen Amperian loop, making the line integral tractable. Without such symmetry, the integral becomes very difficult to evaluate.

The current density is $$J = \frac{I}{\pi(b^2 - a^2)}$$. The current enclosed by radius $$r$$ is $$I_{enc} = J \cdot \pi(r^2 - a^2) = \frac{I(r^2 - a^2)}{b^2 - a^2}$$. Applying Ampère's law: $$B(2\pi r) = \mu_0 I_{enc}$$, giving $$B = \frac{\mu_0 I (r^2 - a^2)}{2\pi r (b^2 - a^2)}$$.

For a finite wire segment, the magnetic field varies both in magnitude and direction around any reasonable Amperian loop, making the line integral $$\oint \vec{B} \cdot d\vec{l}$$ very difficult to evaluate. The Biot-Savart law is more appropriate for such geometries.

The magnetic fields from both solenoids point in the same direction inside the coaxial arrangement. By superposition, the total field is the sum: $$B = \mu_0 n_1 I_1 + \mu_0 n_2 I_2 = \mu_0 \frac{N_1 I_1}{L} + \mu_0 \frac{N_2 I_2}{L} = \mu_0 \frac{(N_1 I_1 + N_2 I_2)}{L}$$.

The current density is $$J = \frac{I}{\pi R^2}$$. The current enclosed by radius $$r$$ is $$I_{enc} = J \cdot \pi r^2 = \frac{I r^2}{R^2}$$. Applying Ampère's law with a circular loop: $$B(2\pi r) = \mu_0 I_{enc}$$, so $$B = \frac{\mu_0 I r}{2\pi R^2}$$.